Download VARIABLES AND COMBINING LIKE TERMS 2.1.1 and 2.1.2 Using

VARIABLES AND COMBINING LIKE TERMS
2.1.1 and 2.1.2
Algebraic expressions can be represented by the perimeters and areas of algebra tiles
y
x
(rectangles and squares) and combinations of algebra
1
tiles. The dimensions of each tile are shown along its
x
x x2
xy
1 1
sides and the area is shown on the tile itself in the
figures at left. When using the tiles, perimeter is the
1
x 1
distance around the exterior of a figure. The area of
x
y
y
2
a figure is the sum of the areas of the individual
y
y
pieces (length times width of each piece).
y
Combining terms that have the same area to write a simpler expression is called
combining like terms. When working without tiles or pictures of the tiles, simply add or
subtract the coefficients of the like terms. The Math Notes boxes on pages 49 and 57
explain the standard order of operations and their use with algebra tiles.
Using Algebra Tiles
Example 1
Example 2
y
x
1
x
xy
x2
x x
x
1
y
2
y
y
y
Example 3
y
y
y2
y2
y
y
1 1 1 1
1 1
y y y 1 1
1
y-2
1 1 1
x2
x2
x2
x
x
x
x
x
x
x
1
1
y
Perimeter = 4x + 4y + 2
Area = x 2 + xy + y 2 + x
Perimeter = 6y + 8
Area = 2y 2 + 3y + 2
Perimeter = 8x + 6
Area = 3x 2 + 7x + 2
Combining Like Terms
Example 4
3x 2 + 7x + 2 + x 2 + 5 + 2x = 4x 2 + 9x + 7
Example 5
x 2 + xy + y 2 + 3xy + y 2 + 7 = x 2 + 4xy + 2y 2 + 7
Example 6:
3x 2 ! 4x + 3 + !x 2 + 3x ! 7 = 2x 2 ! x ! 4
6
Algebra Connections Parent Guide
Problems
Determine the area and the perimeter of each figure.
1.
x2
4.
7.
1 1 1
2.
x x
2
y
5.
y2
y
y
1
2
y
x2
x2
x x
x2
x2
x x
x
x
1 1
x
x
x
x
6.
x
x
9.
x2
x
x
1 1
1 1
y
y
1
1
y
8.
2
3.
x
x2
xy
1
y2
Simplify each expression by combining like terms.
10. 2x 2 + x + 3 + 4x 2 + 3x + 5
11. y 2 + 2y + x 2 + 3y 2 + x 2
12. x 2 ! 3x + 2 + x 2 + 4x ! 7
13. y 2 + 2y ! 3 ! 4y 2 ! 2y + 3
14. 4xy + 3x + 2y ! 7 + 6xy + 2x + 7
15. x 2 ! y 2 + 2x + 3y + x 2 + y 2 + 3y
16. (4x 2 + 4x ! 1) + (x 2 ! x + 7)
17. (y 2 + 3xy + x 2 ) + (2y 2 + 4xy ! x 2 )
18. (7x 2 ! 6x ! 9) ! (9x 2 + 3x ! 4)
19. (3x 2 ! 8x ! 4) ! (5x 2 + x + 1)
Answers
1.
P = 4x + 4
A = x 2 + 2x
2.
P = 4y + 6
A = y2 + 3
3.
P = 4x + 8
A = 4x + 4
4.
P = 4y + 4
A = y 2 + 2y + 1
5.
P = 4y + 4
A = x 2 + 2y
6.
P = 2x + 2y + 6
A = 2x + 2y + 2
7.
P = 8x + 6
A = 4x 2 + 6x + 2
8.
P = 6x
A = x2 + x + 1
9.
P = 6x + 4y
A = 2x 2 + xy + y 2
10.
6x 2 + 4x + 8
11.
4y 2 + 2y + 2x 2
12.
2x 2 + x ! 5
13.
!3y 2
14.
10xy + 5x + 2y
15.
2x 2 + 2x + 6y
16.
5x 2 + 3x + 6
17.
3y 2 + 7xy
18.
!2x 2 ! 9x ! 5
19.
!2x 2 ! 9x ! 5
Chapter 2: Variables and Proportions
7
WRITING AND SIMPLIFYING
ALGEBRAIC EXPRESSIONS
2.1.3 through 2.1.7
An expression mat is an organizing tool that is used to represent algebraic expressions.
Pairs of expression mats can be modified to make an equation mat (see next section). The
upper half of an expression mat is the positive region and the lower half is the negative
region. Positive algebra tiles are shaded and negative tiles are blank. A matching pair of
tiles with one tile shaded and the other one blank represents zero (0).
Tiles may be removed from or moved on an expression mat in one of three ways:
(1) removing the same number of opposite tiles in the same region; (2) flipping a tile
from one region to another. Such moves create “opposites” of the original tile, so a
shaded tile becomes unshaded and an unshaded tile becomes shaded; and (3) removing an
equal number of identical tiles from both the “+” and “-” regions. See the Math Notes
box on page 60.
Examples
+
3x ! 4 can be
represented
various ways.
+
x
x
x
x
x
x
+
x
_
_
+
The expression
mats at right
all represent
zero.
Example 1
x
x
x
x
x
x
x
x
+
= +1
= –1
x
+
x
_
_
Expressions can
be simplified by
moving tiles to
the top (change
the sign) and
looking for
_
zeros.
x
_
3x + 2 ! (2x ! 3)
+
x
x
x
x
x
x
x
x
+
x
= +1
= –1
x
x
3x + 2 ! (2x ! 3)
8
x
_
+
= +1
= –1
_
3x + 2 ! 2x + 3
_
x+5
Algebra Connections Parent Guide
1 ! (2y ! 3) + y ! 2
Example 2
+
+
y
y
+
y
y
y
y
y
= +1
= –1
_
_
1 ! (2y ! 3) + y ! 2
_
1 ! 2y + 3 + y ! 2
!y + 2
Problems
Simplify each expression.
1.
2. +
+
3. +
x
x
x
y
x
_
4.
y
_
+
x
x
x
x
5.
_
6.
+
+
y
y
y
x
_
7.
10.
13.
16.
_
3 + 5x ! 4 ! 7x
4x ! (x + 2)
1 ! 2y ! 2y
!(x + y) + 4x + 2y
8.
11.
14.
17.
y
y
!x ! 4x ! 7
5x ! (!3x + 2)
!3x + 5 + 5x ! 1
3x ! 7 ! (3x ! 7)
_
9.
12.
15.
18.
!(!x + 3)
x ! 5 ! (2 ! x)
3 ! (y + 5)
!(x + 2y + 3) ! 3x + y
Answers
1.
0
2.
2x + 2
3.
2y + 2
4.
!5x + 2
5.
2y ! 1
6.
!y + 5
7.
10.
13.
!2x ! 1
3x ! 2
!4y + 1
8.
11.
14.
!5x ! 7
8x ! 2
2x + 4
9.
12.
15.
x!3
2x ! 7
!y ! 2
16.
3x + y
17.
0
18.
!4x ! y ! 3
Chapter 2: Variables and Proportions
9
SOLVING EQUATIONS
2.1.8 and 2.1.9
Combining two expression mats into an equation mat creates a concrete model for
solving equations. Practicing solving equations using the model will help students
transition to solving equations abstractly with better accuracy and understanding.
In general, and as shown in the first example below, the negative in front of the
parenthesis causes everything inside to “flip” from the top to the bottom or the bottom to
the top of an expression mat, that is, all terms in the expression change signs. After
simplifying the parentheses, simplify each expression mat. Next, isolate the variables on
one side of the equation mat and the non-variables on the other side by removing
matching tiles from both sides. Then determine the value of the variable. Students should
be able to explain their steps. See the Math Notes box on page 69.
Procedure and Example
+
x
x
+
Solve x + 2 ! (!2x) = x + 5 ! (x ! 3) .
x
First build the equation on the equation mat.
_
+
_
x
x
+
x
Second, simplify each side using legal
moves on each expression mat, that is, on
each side of the equation mat.
x
_
+
x
_
x + 2 + 2x = x + 5 ! x + 3
+
x
x
Isolate x-terms on one side and non-x-terms
on the other by removing matching tiles
from both sides of the equation mat.
x
_
+
Finally, since both sides of the equation are
equal, determine the value of x.
_
10
x
x
3x + 2 = 8
x
x
x
_
+
_
3x = 6
x=2
Algebra Connections Parent Guide
Once students understand how to solve equations using an equation mat, they may use
the visual experience of moving tiles to solve equations with variables and numbers. The
procedures for moving variables and numbers in the solving process follow the same
rules. Note: When the process of solving an equation ends with different numbers on
each side of the equal sign (for example, 2 = 4), there is no solution to the problem. When
the result is the same expression or number on each side of the equation (for example,
x + 2 = x + 2 ) it means that all numbers are solutions. See Section 3.2.2-3.2.4 in this
guide.
Example 1
Solve 3x + 3x ! 1 = 4x + 9
3x + 3x ! 1 = 4x + 9
6x ! 1 = 4x + 9
2x = 10
Solution
problem
simplify
add 1, subtract 4x on each side
divide
x=5
Example 2
Solve !2x + 1 ! (!3x + 3) = !4 + (!x ! 2)
Solution
!2x + 1 ! (!3x + 3) = !4 + (!x ! 2)
!2x + 1 + 3x ! 3 = !4 ! x ! 2
x ! 2 = !x ! 6
2x = !4
x = !2
problem
remove parenthesis (flip)
simplify
add x, add 2 to each side
divide
Problems
Solve each equation.
2x ! 3 = !x + 3
1.
4 ! 3x = 2x ! 6
3.
2.
4.
1 + 3x ! x = x ! 4 + 2x
3 + 3x ! (x ! 2) = 3x + 4
6.
!4 + 3x ! 1 = 2x + 1 + 2x
5.
!(x + 3) = 2x ! 6
7.
9.
!x + 3 = 10
4y ! 8 ! 2y = 4
8.
10.
5x ! 3 + 2x = x + 7 + 6x
9 ! (1 ! 3y) = 4 + y ! (3 ! y)
11.
13.
2x ! 7 = !x ! 1
!3x + 7 = x ! 1
12.
14.
!2 ! 3x = x ! 2 ! 4x
1 + 2x ! 4 = !3 ! (!x)
15.
2x ! 1 ! 1 = x ! 3 ! (!5 + x)
16.
!4x ! 3 = x ! 1 ! 5x
17.
10 = x + 6 + 2x
18.
!(x ! 2) = x ! 5 ! 3x
19.
6 ! x ! 3 = 4x ! 8
20.
0.5x ! (!x + 3) = x ! 5
Answers
1.
6.
11.
16.
2
!6
2
no answer
2.
7.
12.
17.
5
!7
3.
8.
all numbers 13.
18.
1 13
Chapter 2: Variables and Proportions
2
no answer
2
!7
4.
9.
14.
19.
1
6
0
2 15
5.
10.
15.
20.
1
!7
2
!4
11
PROPORTIONS
2.2.1 through 2.2.3
Students solve proportional reasoning (ratio) problems in a variety of ways. They may
find the number or cost per one unit and then multiply by the number of units. They may
also organize work in a table. Later in the course students will use ratio equations or
proportions. See Section 5.2.1-5.2.2 for this topic.
Example 1
Mrs. Salem’s student assistant can correct 18 homework papers in 10 minutes. At this rate how
many papers can she correct during a 55 minute class?
18÷10 = 1.8 papers per minute
1.8 x 55 = 99 papers
Example 2
Toby uses 7 tubes of toothpaste every 10 months. How many tubes will he use in 24 months?
How long will 35 tubes last him?
÷10
x 24
Tubes
Months
7
10
0.7
1
16.8
24
x5
÷10
or
7 tubes
=
10 months
so
35 tubes
=
50 months
x 24
x5
about 17 tubes
12
50 months
Algebra Connections Parent Guide
Problems
Solve and explain your reasoning
1.
Alice knows six cups of rice will make enough Spanish rice to feed 15 people. How much
rice is needed to feed 75 people?
2.
Elaine can plant 16 flowers in 10 minutes. How many can she plant in 25 minutes?
3.
Ivanna needs to buy 360 cherries for a large salad. She can buy nine cherries for $0.57.
How much will 360 cherries cost?
4.
A plane travels 3400 miles in eight hours. How far would it travel in six hours at the same
rate?
5.
Leslie can write a 500-word essay in a hour. If she writes an essay in 20 minutes,
approximately how many words should the essay contain?
6.
About eight out of every 100 people in the state have red hair. If a typical classroom in
the state has 25 students, how many would you expect to have red hair? If the typical
middle school has 650 students, how many would you expect to have red hair?
7.
When Carlos rides his bike to school, it takes 15 minutes to go 8 blocks. If he rides at the
same speed, how long should it take him to travel 30 blocks?
8.
Simba the cat is on a diet. Ten pounds of special low-fat food costs $22.50. How much
would 30 pounds cost? How much would 36 pounds cost?
9.
Elizabeth came to bat 110 times in 20 games. How many times should she expect to bat
in 62 games?
10.
Ly can deliver 32 newspapers in 25 minutes on his bike. Next week he needs to deliver
80 newspapers in the same neighborhood. How long should it take him if he works at the
same rate as he did for 32 newspapers?
Answers
1. 30 cups
2. 40 flowers
3. $22.80
4. 2550 miles
6. 2 & 52 people
7. 56.25 min.
8. $67.50; $81.00
9. 341 at-bats
Chapter 2: Variables and Proportions
5. !167 words
10. 62.5 min.
13